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Prove the following identities: (xi) \left( \left( {{\cos }^{3}}B+{{\sin }^{3}}B \right)/\left( \cos B+\sin B \right) \right)+\left( \left( {{\cos }^{3}}B-{{\sin }^{3}}B \right)/\left( \cos B-\sin B \right) \right)=2(xii) (tanB+sinB)/(tanB-sinB)=(secB+1)/(secB-1)
Class 10 | Frank | ICSE | Maths | Trigonometric Identities
Prove the following identities: (xi) \left( \left( {{\cos }^{3}}B+{{\sin }^{3}}B \right)/\left( \cos B+\sin B \right) \right)+\left( \left( {{\cos }^{3}}B-{{\sin }^{3}}B \right)/\left( \cos B-\sin B \right) \right)=2(xii) (tanB+sinB)/(tanB-sinB)=(secB+1)/(secB-1)

(xi)

From the question first we consider Left Hand Side (LHS),

\left( \left( {{\cos }^{3}}B+{{\sin }^{3}}B \right)/\left( \cos B+\sin B \right) \right)+\left( \left( {{\cos }^{3}}B-{{\sin }^{3}}B \right)/\left( \cos B-\sin B \right) \right)

By taking LCM we get,

=\left[ \left( \left( {{\cos }^{3}}B+{{\sin }^{3}}B \right)\left( \cos B-\sin B \right) \right)+\left( \left( {{\cos }^{3}}B-{{\sin }^{3}}B \right)\left( \cos B+\sin B \right) \right) \right]/\left( \left( \cos B+\sin B \right)\left( \cos B-\sin B \right) \right)Then,

=\left( {{\cos }^{4}}B-{{\cos }^{3}}B\sin B+{{\sin }^{3}}B\cos B-{{\sin }^{4}}B+{{\cos }^{4}}B+{{\cos }^{3}}B\sin B-{{\sin }^{3}}B\cos B-{{\sin }^{4}}B \right)/\left( {{\cos }^{2}}B-{{\sin }^{2}}B \right)By simplification we get,

=\left( 2{{\cos }^{4}}B-2{{\sin }^{4}}B \right)/\left( {{\cos }^{2}}B-{{\sin }^{2}}B \right)

Taking common outside,

=2\left( {{\cos }^{4}}B-{{\sin }^{4}}B \right)/\left( {{\cos }^{2}}B-{{\sin }^{2}}B \right)

=\left( 2\left( {{\cos }^{2}}B+{{\sin }^{2}}B \right)\left( {{\cos }^{2}}B-{{\sin }^{2}}B \right) \right)/\left( {{\cos }^{2}}B-{{\sin }^{2}}B \right)

=2\left( {{\cos }^{2}}B+{{\sin }^{2}}B \right)

We know that,{{\sin }^{2}}+{{\cos }^{2}}B=1

=2

Then, Right Hand Side =2

Therefore, LHS = RHS

(xii)

From the question first we consider Left Hand Side (LHS),

=(tanB+sinB)/(tanB-sinB)

We know that, tanB=sinB/cosB

=((sinB/cosB)+sinB)/((sinB/cosB)-sinB)

=(sinB+sinBcosB)/(sinB-sinBcosB)

=(sinB(1+cosB))/(sinB(1-cosB))

=(1+cosB)/(1-cosB)

Also we know that, cosB=1/secB

=(1+(1/cosB))/(1-(1/secB))

=((secB+1)/secB)/((secB-1)secB)

=(secB+1)/(secB-1)

Then, Right Hand Side =(secB+1)/(secB-1)

Therefore, LHS = RHS

i

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